You are watching: A bullet fired from a rifle begins to fall

If you didn't catch the latest MythBusters (yeah! new episodes), they did something straight from the physics textbooks. Just about every text has this example of shooting a bullet horizontally and dropping a bullet from the same height. The idea is that they should hit the ground at the same time. No one but the MythBusters could actually show this demo with a real gun.

I am going to do some calculations, but I want to first write about the physics that accompanies this idea (and you can actually do it your self without the gun). What physics principle does this demo show? Well, it shows two things. First, it shows that the horizontal and vertical motion in projectile motion are independent. That is, what happens in the y-direction, stays in the y-direction. Really, you can treat projectile motion as two separate problems that just happen to take the same amount of time. The second thing it shows is that the velocity in one direction does not effect the velocity in the other direction (which really is the same as the first thing). Let me draw two diagrams, one for a dropped ball and one for ball shot horizontally.

There are some really important things to note in this diagram. First, the force is only in the y-direction. This means that the red ball's x-motion does not change and neither does the blue ball's. Second, the acceleration in the y-direction for both balls will be the same. Even if the balls had different masses, this would be true since a greater mass would have a greater gravitational force from the Earth.

Since the x- and y-motions are independent, and since the balls have the same y-motion they would hit the ground at the same time.

But what about air resistance?

Yes, this complicates things somewhat. First, I will use a typical model for air resistance that depends on some stuff and the speed of the object. Here is the model:

This is the magnitude of the air resistance. The direction is in the opposite direction as the way the object moves. So, here is a bullet shot horizontally from a gun with air resistance.

Here there is a difference. Now there is a component of force that depends on velocity in the y-direction. Will this make it hit the ground at a different time than one that is just dropped? Even if that one also has air resistance? Let's just do it. Of course, I am not the MythBusters. I don't even own a gun. I do own a computer.

Modeling Bullet with air resistance

It seems like I have been using my bullet model a lot lately. Each time I change it just a little. In this case, I am going to start with a .45 caliber bullet like the mythbusters did. They said they used it because it was heavy and slow. That is fine with me. Slow means that I can just use one value for the drag coefficient instead of my super-sonic coefficient table. If you want my python program, I will add it at the end. But here is my first run. I am going to shoot a .45 caliber bullet aimed horizontally 36 inches above the ground. Just as a comparison, I will plot the trajectory of the .45 and another object shot with it but without air resistance.

This gives a range of 102 meters (or feet) which is similar to what the MythBusters found. Notice that without air resistance, the object only goes a few (7 ish) meters more. Now how about a comparison of their vertical motions.

I zoomed in to just the end. Here you can see that the bullet with air resistance hit the ground after 0.44 seconds. The bullet without air resistance hit the ground after 0.431 seconds. Ok, now to add air resistance to the dropped bullet and just drop it (don't shoot it). Also, I will add a plot of a falling bullet without air resistance.

They don't hit at the same time. Actually, all three objects will hit differently. The no air drag object will hit first, then the dropped bullet and then the fired bullet. The above graph looks like just two lines because the dropped bullet and the no drag bullet are so similar. Let me zoom in a bit on that graph.

What if I increase the air resistance? I could redo this with something like a baseball, but I will just increase the coefficient C. Here is what I get.

This is with a ginourmous drag coefficient of C = 20. So, is this right? Should the dropped bullet really hit the ground first with air resistance? Let me draw two force diagrams for a dropped bullet and a fired bullet just a short moment after they are released.

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When initially released, both bullets had the same force in the y-direction. The fired bullet was moving horizontally so that the air resistance was only in the x-direction. So, they both initially had the same vertical acceleration and after a short time about the same velocity (I will call this v2y). But, how do the vertical air resistances compare? Let me call the initial launch speed of the bullet vbullet. One more thing, I am going to write the air resistance as Kv2 just because it is shorter. The air resistance on the dropped bullet will be:

Now let me look at the y-component of air resistance force on the fired bullet. Let me just say that it really hasn't slowed down much (doesn't really matter anyway) in the x-direction. The magnitude of the whole air resistance force would be:

I left off that tiny little y-component of velocity because it is just too small to matter. But what I want is the vertical component of the air resistance. To do this, I need the angle below the horizontal the bullet is traveling. This diagram may be useful:

So, I can write the following:

And then the y-component of the air resistance would be:

There you have it. The y-component of air resistance for the fired bullet still depends on the fired speed of the bullet (since it is proportional to v2). A fired bullet (with air resistance) does not hit the ground at the same time as a dropped bullet. That being said, it was still an awesome demo. The .45 caliber was a great choice since it is slow and heavy and air resistance doesn't play that big of a role.